College Physics by Openstax Chapter 8 Problem 4

(a) What is the momentum of a garbage truck that is 1.20×104 kg and is moving at 30.0 m/s? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum p\textbf{p} is defined as

p=mv,\textbf{p}=m \textbf{v},

where mm is the mass of the system and v\textbf{v} is its velocity.

Part A. The momentum of the garbage truck

The garbage truck has the following quantities given:

mtruck=1.20×104 kg vtruck=30.0 m/s\begin{align*} m_{\text{truck}} & = 1.20 \times 10^{4}\ \text{kg} \ \\ \textbf{v}_{\text{truck}} & = 30.0\ \text{m}/\text{s} \end{align*}

Substitute these values in the formula of momentum to solve for the momentum of the truck.

p=mvp=(1.20×104 kg)(30.0 m/s)p=360000 kg⋅m/sp=3.60×105 kg⋅m/s  (Answer)\begin{align*} \textbf{p} & = m \textbf{v} \\ \textbf{p} & = \left( 1.20 \times 10^{4}\ \text{kg} \right) \left( 30.0\ \text{m}/\text{s} \right) \\ \textbf{p} & = 360000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p} & = 3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. The Speed of a Trash to have the same Momentum as the Truck

The trash has the following properties:

m=8.00 kgp=3.60×105 kg⋅m/s\begin{align*} m & = 8.00\ \text{kg} \\ \textbf{p} & = 3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

We now use the formula for momentum to solve for the unknown velocity.

p=mvv=pmv=3.60×105 kg⋅m/s8.00 kgv=45000 m/sv=4.5×104 m/s  (Answer)\begin{align*} \textbf{p} & = m \textbf{v} \\ \textbf{v} & = \frac{\textbf{p}}{m} \\ \textbf{v} & = \frac{3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s}}{8.00\ \text{kg}} \\ \textbf{v} & = 45000\ \text{m}/\text{s} \\ \textbf{v} & = 4.5 \times 10^{4}\ \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}